# How do you use the quadratic formula to solve x^2=5-x?

Jul 27, 2016

$x = 1.79 , x = - 2.79$

#### Explanation:

First, we need to get all the terms of the equations on one side of the equation
I added x to the left side
$x + {x}^{2} = 5$
then I subtracted 5
$- 5 + x + {x}^{2} = 0$
now I will rearrange the equation to make it look similar to $a {x}^{2} + b x + c = 0$
${x}^{2} + x - 5 = 0$

So we want to use the quadratic formula to find what the roots of x are
the quadratic formula is
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 \cdot a}$

from the equation we will identify what $a , b , \mathmr{and} c$ are
$a = 1$
$b = 1$
$c = - 5$

now we plug all this in to the formula
$x = \frac{- \left(1\right) \pm \sqrt{{\left(1\right)}^{2} - 4 \left(1\right) \left(- 5\right)}}{2 \cdot \left(1\right)}$

next we simplify
$x = \frac{- \left(1\right) \pm \sqrt{1 + 20}}{2}$

$x = \frac{- 1 \pm \sqrt{21}}{2}$

$x = \frac{- 1 \pm 4.58}{2}$

there will be two answers for x because of the $\pm$
$x = \frac{- 1 + 4.58}{2} = \frac{3.58}{2} = 1.79$
$x = \frac{- 1 - 4.58}{2} = - \frac{5.58}{2} = - 2.79$